1959 IMO Problems/Problem 6

Problem

Two planes, $P$ and $Q$, intersect along the line $p$. The point $A$ is in the plane $P$, and the point ${C}$ is in the plane $Q$; neither of these points lies on the straight line $p$. Construct an isosceles trapezoid $ABCD$ (with $AB$ parallel to $DC$) in which a circle can be constructed, and with vertices $B$ and $D$ lying in the planes $P$ and $Q$, respectively.

Solution

We first observe that we must have both lines $AB$ (which we shall denote $a$) and $DC$ (which we shall denote $c$) parallel to $p$, since if one of them is not, then neither can be and they must both intersect $p$ (since they are both coplanar with $p$), making them skew.

Now we note since a circle can be inscribed in the trapezoid, we must have $AB + DC = AD + BC$, and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the average of the lengths of the bases.

We can find this average by dropping perpendicular $AA'$ to $c$ such that $A'$ is on $c$. The average will be $A'C$, which is one of the sides of the rectangle with sides on $a$ and $c$ with vertices at $A$ and ${C}$.

We now draw a circle with center ${C}$ that contains $A'$. The intersections of this circle with $a$ are the two possible values of $B$, from either of which it is trivial to determine the corresponding location for $D$. It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all. Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.


See Also

1959 IMO (Problems) • Resources)
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